The integral of the function \( \frac{1}{\sqrt{x^4+x^2+1}} \) from 1 to infinity is convergent. The comparison test shows that \( \frac{1}{\sqrt{x^4}} = \frac{1}{x^2} \) is convergent over the same interval. Although the antiderivative of the original function is challenging to find, completing the square leads to the integral \( \int \frac{dx}{((x^{2}+\frac{1}{2})^{2}+\frac{3}{4})^{\frac{1}{2}}} \), which suggests the use of trigonometric substitution for evaluation.
PREREQUISITESStudents and educators in calculus, particularly those focusing on integral convergence and evaluation techniques. This discussion is beneficial for anyone tackling complex integrals and seeking to enhance their problem-solving skills in advanced mathematics.
Don't forget the dx.Neil21 said:Ok, so I complete the square and got \int \frac{1}{((x^{2}+\frac{1}{2})^{2}+\frac{3}{4})^{\frac{1}{2}}} . I am still not seeing a way find an antiderivative though.
What I first saw was rewriting the expression as $${\frac{1}{(x^4+x^2+1)^{1/2}}}={\frac{1}{((x^2+({\frac{1}{2}))^2}+(\frac{\sqrt{3}}{2})^2)^{1/2}}}$$.Mark44 said:Don't forget the dx.
$$\int \frac{dx}{((x^{2}+\frac{1}{2})^{2}+\frac{3}{4})^{\frac{1}{2}}}$$
I haven't worked this through, yet, but here's what I would do, FWIW.
Let u = x2 + 1/2
So ##x = \sqrt{u - 1/2}##
Then du = 2xdx
So ##dx = \frac{du}{2x} = \frac{du}{2\sqrt{u - 1/2}}##
After making the substitution and getting everything in terms of u and du, I would try for a trig substitution. I think that might work.
Ooops. Missed that. Sorry guys.Mark44 said:@chet, I think he already established that in the opening post.